Physics Circle home Student Login

Home›Chapters›Heat Transfer and Radiation PYQs›Heat Conduction›Question 44

JEE Main 2019 (10 Jan, Shift 1) · Heat Conduction

Heat Conduction question from JEE Main 2019 (10 Jan, Shift 1)

Q44JEE Main 2019 · 10 JAN · SHIFT 1Easy

Heat Conduction

A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 WK–1m–1, the energy flux through it in the steady state is -

Show answer and solution

Answer: (d)

  1. Temperatures are 103 K10^3\ \text{K} and 102 K10^2\ \text{K}, so ΔT=900 K\Delta T = 900\ \text{K}.
  2. Energy flux HA=KΔTd=0.1×9001\frac{H}{A} = \frac{K\Delta T}{d} = \frac{0.1 \times 900}{1}
  3. =90 W m−2= 90\ \text{W m}^{-2}, so the answer is (d).
Report a mistake

More questions on Heat Conduction